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<p>本题解主要思路取自官方题解，记录下学习过程</p>
</blockquote>
<p>Alice 和 Bob 再次设计了一款新的石子游戏。现有一行 n 个石子，每个石子都有一个关联的数字表示它的价值。给你一个整数数组 stones ，其中 stones[i] 是第 i 个石子的价值。</p>
<p>Alice 和 Bob 轮流进行自己的回合，<strong>Alice 先手</strong>。每一回合，玩家需要从 stones 中移除任一石子，规则如下：</p>
<ul>
<li>如果玩家移除石子后，导致 <strong>所有已移除石子</strong> 的价值 <strong>总和</strong> 可以被 3 整除，那么该玩家就 输掉游戏 。</li>
<li>如果不满足上一条，且移除后没有任何剩余的石子，那么 Bob 将会直接获胜（即便是在 Alice 的回合）。</li>
</ul>
<p>假设两位玩家均采用 <strong>最佳</strong> 决策。如果 Alice 获胜，返回 true ；如果 Bob 获胜，返回 false 。</p>
<p>题目示例：</p>
<img src="https://p6-juejin.byteimg.com/tos-cn-i-k3u1fbpfcp/ca2be621682b41ffae551f7d32a2a137~tplv-k3u1fbpfcp-watermark.image" style="zoom: 80%;">

<h3 id="2、题解"><a href="#2、题解" class="headerlink" title="2、题解"></a>2、题解</h3><h4 id="Step1-石子分类"><a href="#Step1-石子分类" class="headerlink" title="Step1 石子分类"></a>Step1 石子分类</h4><p>根据题意石子分三类，数量余3为0的称为类型0，余3为1的称为类型1，余3为2的称为类型2。</p>
<p>由于在题目规则上进行游戏时需要保证移除石子和不整除3，因此类型0的选取实际上是对总量无影响的，但是如果类型0出现次数为奇数，回合会交换；如果为偶数则回合交换会抵消，判断时可忽略。</p>
<h4 id="Step2-枚举回合信息"><a href="#Step2-枚举回合信息" class="headerlink" title="Step2 枚举回合信息"></a>Step2 枚举回合信息</h4><p>这里我们暂时忽略类型0的选择，只枚举类型1和类型2；</p>
<p>当Alice第一回合选择类型1，Bob只能选择类型1，接着Alice只能选择类型2，Bob只能选择类型1：</p>
<p>1️⃣11 21 21 21…… </p>
<p>如果Alice第一回合选择类型2，Bob只能选择类型2，接着Alice只能选择类型1，Bob只能选择类型2：</p>
<p>2️⃣22 12 12 12……</p>
<p>这里可以发现，实际上Alice第一回合选择之后就奠定了Bob整局选取石子的类型了，那么如果Alice想每一次都赢，那么就必须在过程中让Bob不得已选择另一个类型的石子；</p>
<p>由此对于情况1️⃣来说：</p>
<p>①第一回合Bob如果没有1可选，那么直接在第一回合就输了，此时需保证<strong>类型1有且仅有1个</strong>；</p>
<p>②第一回合平手，此时1的数量是2，第二回合Alice选择了2，此时如果Bob只有2可选也输了，后面的回合都是这种推演，这意味着<strong>类型2的数量应该大于或者等于类型1的数量</strong>；</p>
<p>对于情况2️⃣来说：</p>
<p>①第一回合Bob如果没有2可选，那么直接在第一回合就输了，此时需保证<strong>类型2有且仅有1个</strong>；</p>
<p>②第一回合平手，此时2的数量是2，第二回合Alice选择了1，此时如果Bob只有1可选也输了，后面的回合都是这种推演，这意味着<strong>类型1的数量应该大于或者等于类型2的数量</strong>；</p>
<p>综合以上两种情况，只要<strong>类型1和类型2的石子数量都大于等于1</strong>，那么Alice总可以找到一个方案去赢。</p>
<h4 id="Step3-考虑回合交换"><a href="#Step3-考虑回合交换" class="headerlink" title="Step3 考虑回合交换"></a>Step3 考虑回合交换</h4><p>我们还需要考虑一下类型0为奇数时导致的回合交换，这会使得玩家选取石子的类型发生变化，但由于石子类型为偶数的情况不会导致回合交换，因此我们可以直接假设类型0的数量为1，来看几个例子：</p>
<p><strong>1）当类型0的石子出现在第一回合</strong>，有四种可能性，分别为：</p>
<p>① 01 12 12……</p>
<p>② 10 12 12……</p>
<p>③ 02 21 21……</p>
<p>④ 20 21 21……</p>
<p>在①和②情况下，如果Alice想赢得游戏，就需要出现01 12 11或者10 12 11 等等，也就是需要类型1的数量至少比类型2多3个，即是 <strong>类型1-类型2 &gt;= 3</strong>；</p>
<p>在③和④情况下，如果Alice想赢得游戏，就需要出现02 21 22或者20 21 22 等等，也就是需要类型2的数量至少比类型1多3个，即是 <strong>类型2-类型1 &gt;= 3</strong>；</p>
<p><strong>2）当类型0的石子出现在第一回合之外</strong>，有四种可能性，分别为：</p>
<p>① 11 02 12 12……</p>
<p>② 11 20 12 12……</p>
<p>③ 22 01 21 21……</p>
<p>④ 22 02 21 21……</p>
<p>我们可以发现实际上这和1）的情况是一致的，不过是类型0的位置变换了，回合数增加了，但Alice赢得游戏所需的条件仍是相同的；</p>
<p>因此我们可以得出结论：类型0石子为奇数时，Alice获胜的条件为<strong>类型1的数量至少比类型2多三个</strong> 或者 <strong>类型2的数量至少比类型1多三个</strong>。</p>
<h4 id="Step4-总结"><a href="#Step4-总结" class="headerlink" title="Step4 总结"></a>Step4 总结</h4><p>类型0数量为偶数时：类型1和类型2的石子数量都大于等于1；</p>
<p>类型0数量为奇数时：类型1的数量至少比类型2多三个 或者 类型2的数量至少比类型1多三个；</p>
<h3 id="3、参考代码"><a href="#3、参考代码" class="headerlink" title="3、参考代码"></a>3、参考代码</h3><figure class="highlight java"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br></pre></td><td class="code"><pre><span class="line"><span class="class"><span class="keyword">class</span> <span class="title">Solution</span> </span>&#123;</span><br><span class="line">    <span class="function"><span class="keyword">public</span> <span class="keyword">boolean</span> <span class="title">stoneGameIX</span><span class="params">(<span class="keyword">int</span>[] stones)</span> </span>&#123;</span><br><span class="line">        <span class="keyword">int</span> stone0 = <span class="number">0</span>;</span><br><span class="line">        <span class="keyword">int</span> stone1 = <span class="number">0</span>;</span><br><span class="line">        <span class="keyword">int</span> stone2 = <span class="number">0</span>;</span><br><span class="line">        <span class="keyword">for</span> (<span class="keyword">int</span> stone : stones) &#123;</span><br><span class="line">            <span class="keyword">if</span> (stone % <span class="number">3</span> == <span class="number">0</span>) &#123;</span><br><span class="line">                stone0++;</span><br><span class="line">            &#125; <span class="keyword">else</span> <span class="keyword">if</span> (stone % <span class="number">3</span> ==<span class="number">1</span>) &#123;</span><br><span class="line">                stone1++;</span><br><span class="line">            &#125; <span class="keyword">else</span> &#123;</span><br><span class="line">                stone2++;</span><br><span class="line">            &#125;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">if</span> (stone0 % <span class="number">2</span> == <span class="number">0</span>) &#123;</span><br><span class="line">            <span class="keyword">return</span> stone1 &gt;= <span class="number">1</span> &amp;&amp; stone2 &gt;= <span class="number">1</span>;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">return</span> stone1 - stone2 &gt;= <span class="number">3</span> || stone2 - stone1 &gt;= <span class="number">3</span>;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure></article><div class="post-copyright"><div class="post-copyright__author"><span class="post-copyright-meta">文章作者: </span><span class="post-copyright-info"><a href="mailto:undefined">h0ss</a></span></div><div class="post-copyright__type"><span class="post-copyright-meta">文章链接: </span><span class="post-copyright-info"><a href="https://blog.gpnusz.cn/2022/01/21/%E3%80%90%E5%8A%9B%E6%89%A3%E9%A2%98%E8%A7%A3%E3%80%91%E7%9F%B3%E5%AD%90%E6%B8%B8%E6%88%8F-IX/">https://blog.gpnusz.cn/2022/01/21/%E3%80%90%E5%8A%9B%E6%89%A3%E9%A2%98%E8%A7%A3%E3%80%91%E7%9F%B3%E5%AD%90%E6%B8%B8%E6%88%8F-IX/</a></span></div><div class="post-copyright__notice"><span class="post-copyright-meta">版权声明: </span><span class="post-copyright-info">本博客所有文章除特别声明外，均采用 <a href="https://creativecommons.org/licenses/by-nc-sa/4.0/" target="_blank">CC BY-NC-SA 4.0</a> 许可协议。转载请注明来自 <a href="https://blog.gpnusz.cn" target="_blank">后端学习记录</a>！</span></div></div><div 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